This is not a complete answer but hopefully shows why this may be true. Let's assume $a_i=a_i^6$ then our inequality becomes $$
g(n)=\sum _{i=1}^n \left(a_i^3 a_{i+1}^3-a_i^2 a_{i+1}^2 a_{i+2}^2\right)\geq0
$$ where the indexes are cyclical.
Let's assume $n\leq6$ and let's make the substitution $a_i=x_1+...+x_i$ for each $1\leq i\leq n$.
Simplifying (using Mathematica) gives a polynomial with only one negative coef, the one before $x_1^4 x_2 x_n$ and this coef is always $-1$.
If someone manages to prove this for all $n$ it may be enough to establish the inequality.
Here is the $n=4$ polynomial for reference.$$
3 x_1^4 x_2^2+10 x_1^3 x_2^3+12 x_1^2 x_2^4+6 x_1 x_2^5+x_2^6+7 x_1^4 x_2 x_3+29 x_1^3 x_2^2 x_3+42 x_1^2 x_2^3 x_3+25 x_1 x_2^4 x_3+5 x_2^5 x_3+7 x_1^4 x_3^2+35 x_1^3 x_2 x_3^2+64 x_1^2 x_2^2 x_3^2+48 x_1 x_2^3 x_3^2+12 x_2^4 x_3^2+14 x_1^3 x_3^3+47 x_1^2 x_2 x_3^3+51 x_1 x_2^2 x_3^3+17 x_2^3 x_3^3+13 x_1^2 x_3^4+28 x_1 x_2 x_3^4+14 x_2^2 x_3^4+6 x_1 x_3^5+6 x_2 x_3^5+x_3^6-x_1^4 x_2 x_4+x_1^3 x_2^2 x_4+6 x_1^2 x_2^3 x_4+5 x_1 x_2^4 x_4+x_2^5 x_4+7 x_1^4 x_3 x_4+26 x_1^3 x_2 x_3 x_4+46 x_1^2 x_2^2 x_3 x_4+36 x_1 x_2^3 x_3 x_4+9 x_2^4 x_3 x_4+21 x_1^3 x_3^2 x_4+66 x_1^2 x_2 x_3^2 x_4+72 x_1 x_2^2 x_3^2 x_4+24 x_2^3 x_3^2 x_4+26 x_1^2 x_3^3 x_4+56 x_1 x_2 x_3^3 x_4+28 x_2^2 x_3^3 x_4+15 x_1 x_3^4 x_4+15 x_2 x_3^4 x_4+3 x_3^5 x_4+3 x_1^4 x_4^2+7 x_1^3 x_2 x_4^2+10 x_1^2 x_2^2 x_4^2+8 x_1 x_2^3 x_4^2+2 x_2^4 x_4^2+11 x_1^3 x_3 x_4^2+28 x_1^2 x_2 x_3 x_4^2+30 x_1 x_2^2 x_3 x_4^2+10 x_2^3 x_3 x_4^2+16 x_1^2 x_3^2 x_4^2+34 x_1 x_2 x_3^2 x_4^2+17 x_2^2 x_3^2 x_4^2+12 x_1 x_3^3 x_4^2+12 x_2 x_3^3 x_4^2+3 x_3^4 x_4^2+2 x_1^3 x_4^3+3 x_1^2 x_2 x_4^3+3 x_1 x_2^2 x_4^3+x_2^3 x_4^3+3 x_1^2 x_3 x_4^3+6 x_1 x_2 x_3 x_4^3+3 x_2^2 x_3 x_4^3+3 x_1 x_3^2 x_4^3+3 x_2 x_3^2 x_4^3+x_3^3 x_4^3
$$